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Three-Dimensional Solutions*

4
Three-Dimensional Solutions *

Three-dimensional solutions involve all three spatial coordinates. The discussion of three-dimensional waves in Cartesian coordinates is fairly straightforward and will be discussed next through a cavity example.

4.1Rectangular Cavity with PEC Boundaries: TM Modes

Consider a rectangular box of a × b × h as shown in Figure 4.1.

FIGURE 4.1
Rectangular cavity with PEC boundaries. Positive z-direction is opposite to the direction shown in the figure for the right-handed Cartesian coordinate system.

This cavity can be constructed by taking a piece of a rectangular waveguide of length h and closing the box with end PEC plates at the ends z = 0 and h. For TM modes, the additional boundary conditions are

(4.1)

{\tilde{E}}_{tan} = 0 at z = 0 or h .

${\tilde{E}}_{tan} = 0 at z = 0 or h .$

This boundary condition is not on ${\tilde{E}}_{z}$ ${\tilde{E}}_{z}$ , but Equation 4.1 is equivalent to (see Equation 2.24)

(4.2)

\frac{\partial {\tilde{E}}_{z}}{\partial z} = 0 at z = 0 or h .

$\frac{\partial {\tilde{E}}_{z}}{\partial z} = 0 at z = 0 or h .$

It follows that

(4.3)

\begin{array}{l} {\tilde{E}}_{z} (x, y, z) = E_{m n l} sin \frac{m π x}{a} sin \frac{n π y}{b} cos \frac{l π z}{h}, \\ m = 1, 2, \dots, \infty, n = 1, 2, \dots, \infty, l = 0, 1, \dots, \infty . \end{array}

$\begin{array}{l} {\tilde{E}}_{z} (x, y, z) = E_{m n l} sin \frac{m π x}{a} sin \frac{n π y}{b} cos \frac{l π z}{h}, \\ m = 1, 2, \dots, \infty, n = 1, 2, \dots, \infty, l = 0, 1, \dots, \infty . \end{array}$

In the above, the separation constants are k_x = mπ/a, k_y = nπ/b, and k_z = lπ/h. Thus, we obtain

(4.4)

k^{2} = ω^{2} μ ε = {(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2} .

$k^{2} = ω^{2} μ ε = {(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2} .$

When the size of the cavity is fixed, the frequency ω in Equation 4.4 is dependent on m, n, l, μ, and ε. For a given mode and a specified medium, the frequency ω is a definite value, labeled as the resonant frequency of the cavity. Its value (in Hz) is given by

(4.5)

\begin{array}{l} {(f_{r})}_{{TM}_{m n l}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2}]}^{1 / 2} \\ m = 1, 2, \dots, \infty, n = 1, 2, \dots, \infty, l = 0, 1, \dots, \infty . \end{array}

$\begin{array}{l} {(f_{r})}_{{TM}_{m n l}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2}]}^{1 / 2} \\ m = 1, 2, \dots, \infty, n = 1, 2, \dots, \infty, l = 0, 1, \dots, \infty . \end{array}$

The lowest TM resonant frequency is given by

(4.6)

{(f_{r})}_{{TM}_{110}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{π}{a})}^{2} + {(\frac{π}{b})}^{2}]}^{1 / 2} .

${(f_{r})}_{{TM}_{110}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{π}{a})}^{2} + {(\frac{π}{b})}^{2}]}^{1 / 2} .$

4.2Rectangular Cavity with PEC Boundaries: TE Modes

The additional boundary conditions to be satisfied in this case are

(4.7)

{\tilde{H}}_{z} = 0, z = 0 or h .

${\tilde{H}}_{z} = 0, z = 0 or h .$

The expression for ${\tilde{H}}_{z}$ ${\tilde{H}}_{z}$ is now easily obtained:

(4.8)

\begin{array}{l} {\tilde{H}}_{z} = H_{m n l} cos \frac{m π x}{a} cos \frac{n π y}{b} sin \frac{l π z}{h} \\ m = 0, 1, 2, \dots, \infty, n = 0, 1, 2, \dots, \infty, l = 1, 2, \dots, \infty, \end{array}

$\begin{array}{l} {\tilde{H}}_{z} = H_{m n l} cos \frac{m π x}{a} cos \frac{n π y}{b} sin \frac{l π z}{h} \\ m = 0, 1, 2, \dots, \infty, n = 0, 1, 2, \dots, \infty, l = 1, 2, \dots, \infty, \end{array}$

but m = n = 0 is excluded.

The resonant frequency can now be obtained as

(4.9)

\begin{array}{l} {(f_{r})}_{{TE}_{m n l}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2}]}^{1 / 2} \\ m = 0, 1, 2, \dots, \infty, n = 0, 1, 2, \dots, \infty, l = 1, 2, \dots, \infty, \end{array}

$\begin{array}{l} {(f_{r})}_{{TE}_{m n l}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{m π}{a})}^{2} + {(\frac{n π}{b})}^{2} + {(\frac{l π}{h})}^{2}]}^{1 / 2} \\ m = 0, 1, 2, \dots, \infty, n = 0, 1, 2, \dots, \infty, l = 1, 2, \dots, \infty, \end{array}$

but m = n = 0 is excluded.

If h > a > b, then the lowest TE resonant frequency is given by

(4.10)

{(f_{r})}_{{TE}_{101}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{π}{a})}^{2} + {(\frac{π}{h})}^{2}]}^{1 / 2}, h > a > b .

${(f_{r})}_{{TE}_{101}} = \frac{1}{2 π \sqrt{μ ε}} {[{(\frac{π}{a})}^{2} + {(\frac{π}{h})}^{2}]}^{1 / 2}, h > a > b .$

The resonant frequency given by Equation 4.10 is lower than that given by Equation 4.6 if h > a > b.

4.3Q of a Cavity

The waveguide and cavity problems assumed PEC boundary conditions. After obtaining the fields, we can relax the ideal assumptions by calculating the losses when the walls are not perfect [1]. We illustrate the technique for TE₁₀₁ mode. It is convenient to write the fields in the following forms:

(4.11)

{\tilde{E}}_{y} = E_{0} sin \frac{π x}{a} sin \frac{π z}{h},

${\tilde{E}}_{y} = E_{0} sin \frac{π x}{a} sin \frac{π z}{h},$

(4.12)

{\tilde{H}}_{x} = - j \frac{E_{0}}{η} \frac{λ}{2 h} sin \frac{π x}{a} cos \frac{π z}{h},

${\tilde{H}}_{x} = - j \frac{E_{0}}{η} \frac{λ}{2 h} sin \frac{π x}{a} cos \frac{π z}{h},$

(4.13)

{\tilde{H}}_{z} = j \frac{E_{0}}{η} \frac{λ}{2 a} cos \frac{π x}{a} sin \frac{π z}{h} .

${\tilde{H}}_{z} = j \frac{E_{0}}{η} \frac{λ}{2 a} cos \frac{π x}{a} sin \frac{π z}{h} .$

We obtain the power loss in the walls by calculating the surface currents. The surface currents are obtained from the magnetic fields:

\begin{array}{l} Front : & {\tilde{K}}_{y} & = & - {\tilde{H}}_{x} |_{z = h'} & Back : & {\tilde{K}}_{y} & = & {\tilde{H}}_{x} |_{z = 0'} \\ Left side : & {\tilde{K}}_{y} & = & - {\tilde{H}}_{z} |_{x = 0'} & Right side : & {\tilde{K}}_{y} & = & {\tilde{H}}_{z} |_{x = a'} \\ Top : & {\tilde{K}}_{x} & = & - {\tilde{H}}_{z} |_{y = b'} & Bottom: & {\tilde{K}}_{x} & = & {\tilde{H}}_{z} |_{y = 0'} \\ {\tilde{K}}_{z} & = & - {\tilde{H}}_{x} |_{y = b'} & {\tilde{K}}_{z} & = & - {\tilde{H}}_{x} |_{y = 0'} \end{array}

$\begin{array}{l} Front : & {\tilde{K}}_{y} & = & - {\tilde{H}}_{x} |_{z = h'} & Back : & {\tilde{K}}_{y} & = & {\tilde{H}}_{x} |_{z = 0'} \\ Left side : & {\tilde{K}}_{y} & = & - {\tilde{H}}_{z} |_{x = 0'} & Right side : & {\tilde{K}}_{y} & = & {\tilde{H}}_{z} |_{x = a'} \\ Top : & {\tilde{K}}_{x} & = & - {\tilde{H}}_{z} |_{y = b'} & Bottom: & {\tilde{K}}_{x} & = & {\tilde{H}}_{z} |_{y = 0'} \\ {\tilde{K}}_{z} & = & - {\tilde{H}}_{x} |_{y = b'} & {\tilde{K}}_{z} & = & - {\tilde{H}}_{x} |_{y = 0'} \end{array}$

If the conducting walls have surface resistance R_s, then the losses are given by

(4.14)

W_{L} = \frac{R_{s}}{2} \{2 \int_{0}^{b} \int_{0}^{a} {{|{\tilde{H}}_{x}|}^{2} |}_{z = 0} d x d y + 2 \int_{0}^{h} \int_{0}^{b} {{|{\tilde{H}}_{z}|}^{2} |}_{x = 0} d y d z + 2 \int_{0}^{h} \int_{0}^{a} [{|{\tilde{H}}_{x}|}^{2} + {|{\tilde{H}}_{z}|}^{2}] d x d z\},

$W_{L} = \frac{R_{s}}{2} \{2 \int_{0}^{b} \int_{0}^{a} {{|{\tilde{H}}_{x}|}^{2} |}_{z = 0} d x d y + 2 \int_{0}^{h} \int_{0}^{b} {{|{\tilde{H}}_{z}|}^{2} |}_{x = 0} d y d z + 2 \int_{0}^{h} \int_{0}^{a} [{|{\tilde{H}}_{x}|}^{2} + {|{\tilde{H}}_{z}|}^{2}] d x d z\},$

(4.15)

W_{L} = \frac{R_{s} λ^{2}}{8 η^{2}} E_{0}^{2} [\frac{a b}{h^{2}} + \frac{b h}{a^{2}} + \frac{1}{2} (\frac{a}{h} + \frac{h}{a})] .

$W_{L} = \frac{R_{s} λ^{2}}{8 η^{2}} E_{0}^{2} [\frac{a b}{h^{2}} + \frac{b h}{a^{2}} + \frac{1}{2} (\frac{a}{h} + \frac{h}{a})] .$

If the losses are neglected, then the energy in the cavity passes between electric and magnetic fields, and we may calculate the total energy in the cavity by finding the energy storage at the instant when it is maximum:

(4.16)

U = {(U_{E})}_{MAX} = \frac{ε}{2} \int_{0}^{h} \int_{0}^{b} \int_{0}^{a} {|{\tilde{E}}_{y}|}^{2} d x d y d z = \frac{ε a b h}{8} E_{0}^{2} .

$U = {(U_{E})}_{MAX} = \frac{ε}{2} \int_{0}^{h} \int_{0}^{b} \int_{0}^{a} {|{\tilde{E}}_{y}|}^{2} d x d y d z = \frac{ε a b h}{8} E_{0}^{2} .$

The quality factor Q of a cavity is a quantitative measure of how well the cavity is acting as a resonator. It is defined by

(4.17)

Q = \frac{ω_{r} U}{W_{L}} .

$Q = \frac{ω_{r} U}{W_{L}} .$

From Equations 4.14, 4.16, and 4.17, we obtain

(4.18)

Q = \frac{π η}{4 R_{s}} [\frac{2 b {(a^{2} + h^{2})}^{3 / 2}}{a h (a^{2} + h^{2}) + 2 b (a^{3} + h^{3})}],

$Q = \frac{π η}{4 R_{s}} [\frac{2 b {(a^{2} + h^{2})}^{3 / 2}}{a h (a^{2} + h^{2}) + 2 b (a^{3} + h^{3})}],$

which for a cube reduces to

(4.19)

Q = 0.742 \frac{η}{R_{s}} .

$Q = 0.742 \frac{η}{R_{s}} .$

For an air dielectric η = 377, and for a copper conductor at 10 GHz, R_s ≈ 0.0261, giving a quality factor Q = 10,730. Such a large value of Q cannot be obtained by lumped circuits or even with resonant lines.

Reference

1.Ramo, S., Whinnery, J.R., and Van Duzer, T. V., Fields and Waves in Communication Electronics, 3rd edn., Wiley, New York, 1994.

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Table of Contents for 4. Three-Dimensional Solutions

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4.1Rectangular Cavity with PEC Boundaries: TM Modes

4.2Rectangular Cavity with PEC Boundaries: TE Modes

4.3Q of a Cavity

Reference

Table of Contents for
4. Three-Dimensional Solutions